Theorem 23.4 of Munkres’ Topology

Question

Let $A$ be a connected subspace of $X$. If $A\subseteq B\subseteq \overline{A}$, then $B$ is also connected.

My attempt:

Approach(1): Assume towards contradiction, $B$ is not connected. Then $\exists P,Q\in \mathcal{T}_B$ such that $P,Q\neq \phi$, $P\cap Q=\phi$, and $P\cup Q=B$. Let $P=B\cap R$ and $Q=B\cap S$, where $R,S\in \mathcal{T}_X$. Since $P,Q\neq \phi$, $\exists x\in P=B\cap R$ and $\exists y\in Q=B\cap S$, $x\neq y$, because $P\cap Q=\phi$. Since $x\in R$ and $y\in S$, we have $R\in \mathcal{N}_x$ and $S\in \mathcal{N}_y$. So $x,y\in B\subseteq \overline{A}$. By definition of closure, $A\cap R\neq \phi$ and $A\cap S\neq \phi$. Both $A\cap R$, $A\cap S\in \mathcal{T}_A$. Since $P\cap Q=B\cap (R\cap S)=\phi$, we have $(A\cap R)\cap (A\cap S)=A\cap (R\cap S)\subseteq B\cap (R\cap S)=\phi$, inclusion follows from the hypothesis $A\subseteq B$. Hence $A\cap R$ and $A\cap S$ is disjoint set. $P\cup Q =B\cap (R\cup S)=B$. Which implies $A\subseteq B\subseteq (R\cup S)$. So $(A\cap R)\cup (A\cap S)=A \cap (R\cup S)=A$. Hence union of $A\cap R$ and $A\cap S$ is $A$. Thus $A\cap R$ and $A\cap S$ form a separation of subspace $A$. Which contradicts our initial assumption that $A$ is connected. Is this proof correct?

Approach(2): Assume towards contradiction. Let $P$ and $Q$ be the separation of $B$. By lemma 23.2, $A\subseteq P$ or $A\subseteq Q$. WLOG, assume $A\subseteq P$. Since $P$ is closed in $B$, $(\overline{A})_B$(Closure in $B$)$\subseteq P$, by property of closure set. Note $(\overline{A})_B=\overline{A}\cap B$, $\overline{A}$ denote closure in $X$. Since $B\subseteq \overline{A}$, we have $\overline{A}\cap B=B\subseteq P$. We already know $P\subseteq B$. Thus $B=P$. So $P\cup Q=B\cup Q=B$ which implies $Q=\phi$. Thus we reach contradiction. This proof is slight variation of Munkres’ proof. Is this proof correct?

Approach(3): (Munkres’ Proof) Assume towards contradiction. Let $P$ and $Q$ be the separation of $B$. By lemma 23.2, $A\subseteq P$ or $A\subseteq Q$. WLOG, assume $A\subseteq P$. By exercise 6 section 18(a) (link: Exercise 6, Section 17 of Munkres’ Topology), $B\subseteq \overline{A} \subseteq \overline{P}$. Since $P$ is closed in $B$, $(\overline{P})_B=P=\overline{P} \cap B$. So $P\cap Q= (\overline{P}\cap Q)\cap B$. Since $(\overline{P}\cap Q)\subseteq Q\subseteq B$, we have $P\cap Q= (\overline{P}\cap Q)\cap B=\overline{P}\cap Q=\phi$. Thus $\overline{P}\cap Q=\phi$. $B\cap Q\subseteq \overline{A} \cap Q\subseteq \overline{P}\cap Q=\phi$. Which implies $B\cap Q=\phi \Rightarrow Q=\phi$. Which contradict our initial assumption. Is this explanation correct? Munkres didn’t explicitly go into the subspace details. In this approach, I tried to fill in the details.

At first glance Munkres’ proof looks fine, but it’s not the “complete” proof(IMO), I didn’t realise this thing at first.

Edit: Better version of approach(3) is the following: $B\subseteq \overline{A} \subseteq \overline{P}$. So $B\subseteq \overline{P}$. $B\subseteq \overline{P} \cap B=(\overline{P})_B=P$, since $P$ is closed in $B$. Thus $B\subseteq P$, which implies $Q=\phi$. We reach contradiction.

Answer 1

I'll give my favourite proof: let $f: B \to \{0,1\}$ be continuous (where the codomain is discrete). Then $f\restriction_A$ is continuous and by connectedness of $A$, $f[A]=\{i\}$ for some $i \in \{0,1\}$. As $f$ and the constant function with value $i$ coincide on the dense subset $A$ and $\{0,1\}$ is Hausdorff, we have that $f \equiv i$ and so $B$ is connected. Or use that by continuity $f[B]\subseteq f[\overline{A}] \subseteq \overline{f[A]}=\overline{\{i\}}=\{i\}$ instead, of you prefer.

This uses the standard alternative definition of connectedness of $X$ by saying $X$ is connected iff every continuous $f:X \to \{0,1\}$ must be constant. (Sketch of proof: if it were non-constant, then $f^{-1}[\{0\}$ and $f^{-1}[\{1\}$ would form a separation of $X$ and if $A \cup B$ is a non-trivial separation into two disjoint open sets, we can define $F \equiv 0$ on $A$ and $1$ on $B$ to find a non-constant $f$, as required). I like such proofs better than messing with separations.

Answer 2

Your attempts as of now are notationally dense and a bit of work to follow. I suggest the following approach if you want to use a separation type argument: Show that $B$ is disconnected if and only if there are closed (in B), non-empty, disjoint subsets of $B$ that union to $B$. Using this, if you assume $B$ is not connected, show that $A$ must lie completely in one of the closed (in B) sets which union to $B$. You will be able to find a closed set containing $A$ that is a subset of $\overline{A}$, which contradicts the fact that $\overline{A}$ is the smallest closed subset containing $A$.

Taking some inspiration from Henno Brandsma's post about locally constant functions, we present a proof based on the following characterization of connectedness:

A topological space $X$ is connected if and only if for every open cover $\mathcal{U}$ of $X$ there is a chain in $\mathcal{U}$ between any two points in $X$ (Henno Brandsma defines these terms in the post linked above).

Proof (that $A\subseteq B\subseteq \overline{A}$ and $A$ connected implies $B$ connected):

Let $\mathcal{U}$ be an open cover of $B$ and fix $x,y\in B$. There are four cases:

Case 1: $x$ and $y$ are in $A$. In this case, there is a chain in $\mathcal{U}$ between $x$ and $y$ as $A$ is connected.

Case 2: $x$ is in $A$ and $y$ is not in $A$. In this case, $y$ is a limit point of $A$. Since $\mathcal{U}$ is an open cover of $B$, then there exists an open set $U\in\mathcal{U}$ containing $y$. As $y$ is a limit point of $A$, then there exists a $z\in U\cap (A\setminus\{y\})$. By connectedness of $A$, there is a chain in $\mathcal{U}$ from $x$ to $z$. Link $U$ to the end of the chain from $x$ to $z$ to get a chain from $x$ to $y$ in $\mathcal{U}$.

Case 3: $x$ is not in $A$ and $y$ is in $A$. This case is similar to Case 2.

Case 4: $x$ and $y$ are not in $A$. I will leave this as an exercise to you.

This completes the proof that $B$ is connected.

Answer 3

All approaches in your answer seem to be correct, but to be honest: They get bogged down in details and are too complicated. Henno Brandsma's proof is most elegant and transparent, but let me nevertheless give a short proof based on separation. It is similar to approach 2.

So let $P, Q$ be a separation of $B$. By lemma 23.2, $A\subseteq P$ or $A\subseteq Q$. WLOG, assume $A\subseteq P$. Noting $(\overline{A})_B=\overline{A}\cap B = B$, we get $B \subseteq P$ since $P = B \setminus Q$ is closed in $B$. Since $P \subseteq B$, we see that $P = B$. This contradicts the assumption that $P, Q$ is a separation of $B$.